Question

If $\alpha$ and $\beta$ are roots of $x^2+px+q=0$ then find value of ${\alpha }^4+{\beta }^4$ in terms of $p$ and $q$

Answer 1

$x^2+px+q=0$

Here, $a=1,b=p,c=q$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-p}{1}=-p$ ----- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{q}{1}=q$ ----- ( 2 )

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ $(-p{)}^2={\alpha }^2+{\beta }^2+2(q)$ [ From ( 1 ) and ( 2 )]

$\therefore$ ${\alpha }^2+{\beta }^2=p^2-2q$ ----- ( 3 )

$\Rightarrow$ $(\alpha +\beta {)}^4={\alpha }^4+{\beta }^4+4{\alpha }^3\beta +4\alpha {\beta }^3+6{\alpha }^2+{\beta }^2$

$\Rightarrow$ $(\alpha +\beta {)}^4={\alpha }^4+{\beta }^4+4\alpha \beta ({\alpha }^2+{\beta }^2)+6{\alpha }^2{\beta }^2$

Now from ( 1 ), ( 2 ) and ( 3 )

$\Rightarrow$ $(-p{)}^4={\alpha }^4+{\beta }^4+4q(p^2-2q)+6(q{)}^2$

$\Rightarrow$ $p^4={\alpha }^4+{\beta }^4+4p^{2}q-8q^2+6q^2$

$\Rightarrow$ $p^4={\alpha }^4+{\beta }^4+4p^{2}q-2q^2$

$\Rightarrow$ ${\alpha }^4+{\beta }^4=p^4-4p^{2}q+2q^2$

$\therefore$ ${\alpha }^4+{\beta }^4=p^2(p^2-4q)+2q^2$